Question

solve the ode
y''-2y' = 2e^(2t)

Answer 1

Do you know how to find the complimentary solution to this?

Answer 2

No...

Answer 3

\(\sf \color{}{r^2-2r=0 \Rightarrow r(r-2)}\) so your \(\sf \color{}{y_c = c_1+c_2e^{2t}}\)

Answer 4

To find your general solution, now you must find your particular solution.

Answer 5

I suggest using method of undetermined coefficients, are you familiar with that method?

Answer 6

Or annihilator method. Whichever you are more comfortable using.

Answer 7

Remember that \(\sf \color{red}{y_g = y_c + y_p}\)

Answer 8

would it be too much to ask you to walk through one of those methods with me? I'm sort of drowning in this class right now..

Answer 9

If you're drowning, I think it might be best if you did it yourself and I helped guide you. This is something that's really important in ODE's and you will be tested on.

Answer 10

Will that work for you?

Answer 11

yeah, that sounds great!

Answer 12

Ok, so are you familiar with undetermined coefficients or annihilator method for this?

Answer 13

Or which one have you been taught in class?

Answer 14

we've been taught the undetermined method. ive never even heard of the anihalator method or whatever

Answer 15

Ok, well you know that for undetermined coefficients, you are taking sort of "a guess" right? For \(e^{t}\) Yes?

Answer 16

For your guess, you're going to use: \(\sf \color{blue}{Ae^{2t}}\)
Now, using your complimentary solution that you have above: \(\sf \color{red}{y_c}\)
find: \(\sf \color{green}{y_c'~and~y_c''}\)

Answer 17

Can you do that?

Answer 18

What you're doing is trying to make the e\(^{2t}\) disappear by doing this.

Answer 19

yeah, that sounds about right. im trying to follow along using my notes here, sorry im replying slowly.